Mechanics Underkill

Classical Mechanics Level 4

Two blocks $A$ and $B$ of equal masses are sliding down along parallel lines on a plane inclined at an angle of $\theta =45^\circ$ with the horizontal, as shown in the figure. The coefficients of kinetic friction are $0.2$ and $0.3$ for $A$ and $B,$ respectively. At $t=0,$ both the blocks are at rest, and block $A$ is $2$ meters behind block $B$ .

Find the time (in seconds) and distance (in meters) from the initial position of $A$ to where the front faces of the blocks come in line on the inclined plane, as illustrated in the figure. $\big($ Use $g=\SI[per-mode=symbol]{10}{\meter\per\second\squared}.\big)$

Input your answer as the sum of the answers for the time and distance, to 3 decimals.

Source : JEE 2004

The answer is 18.378.

1 solution

Aditya Narayan Sharma
Apr 17, 2017

Relevant wiki: Motion Along Inclined Planes

Suppose A and B meet at M ,and AM $= x$ with the time taken for both of these to meet is $T$

Acceleration in the direction of falling is $\displaystyle a=g(\sin \theta-\mu\cos \theta)=\dfrac{g}{\sqrt{2}}\left(1-\mu\right)$

Displacement of A is $x$ while of B is $x-2$

For A , $\displaystyle x=\dfrac{1}{2}aT^2 = \dfrac{5}{\sqrt{2}}T^2\left(1-\mu_A \right)$

For B , $\displaystyle x-2=\dfrac{1}{2}aT^2 = \dfrac{5}{\sqrt{2}}T^2\left(1-\mu_B \right)$

Subtracting the equations we have,

$\displaystyle \dfrac{5}{\sqrt{2}}T^2\left(\mu_B-\mu_A \right)=2$ which upon solving gives $\displaystyle T=2^{5/4}$

Substituting T we get $x=16$

The required answer is therefore $\displaystyle \boxed{x+T\approx 18.37}$

Mechanics Underkill

The answer is 18.378.

1 solution

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